『NYIST』第八届河南省ACM竞赛训练赛[正式赛一] E题

E - ECrawling in process...Crawling failedTime Limit:1000MSMemory Limit:262144KB 64bit IO Format:%I64d & %I64uSubmitStatusPracticeCodeForces 237CDescriptionYou've decided to carry out a survey in the theory of prime numbers. Let us remind you that a prime number is a positive integer that has exactly two distinct positive integer divisors.Consider positive integers a,a + 1, ...,b(a ≤ b). You want to find the minimum integerl(1 ≤ l ≤ b - a + 1) such that for any integerx(a ≤ x ≤ b - l + 1) amongl integers x,x + 1, ...,x + l - 1 there are at least k prime numbers.Find and print the required minimum l. If no valuel meets the described limitations, print -1.InputA single line contains three space-separated integers a, b, k (1 ≤ a, b, k ≤ 106; a ≤ b).OutputIn a single line print a single integer — the required minimuml. If there's no solution, print -1.Sample InputInput

2 4 2

Output

3

Input

6 13 1

Output

4

Input

1 4 3

Output

-1

#include<algorithm>#include<stdio.h>#include<string.h>using namespace std;#define maxn 1000005int p[maxn]= {0}; //前i位有sum[i]个素数int d[maxn];//是素数标记为1，否则标记为0int a,b,k;void shaifa(){int i,j;d[0]=d[1]=1;for(i=2; i<maxn; i++){p[i]=p[i-1];if(!d[i]){p[i]++;for(j=2*i; j<maxn; j+=i)//或者:for(j=1;i*j<maxn;j++){d[i*j]=1;}{d[j]=1;}}}}bool check(int x)//传入二分查找的中间值{for(int i=a; i<=b-x+1; i++)if(p[i+x-1]-p[i-1]<k)//此区间的素数小于kreturn 0;return 1;//满足条件返回1}int main(){shaifa();while(~scanf("%d%d%d",&a,&b,&k)){if(p[b]-p[a-1]<k)//取极限，如果不满足条件{printf("-1\n");return 0;}int left=1;//左边界int right=b-a+1;//右边界int num;while(left<=right)//二分查找{int mid=(left+right)/2;//中间值if(check(mid))//满足条件{num=mid;right=mid-1;//缩小范围继续查找}elseleft=mid+1;//改变左边界}printf("%d\n",num);}}

[/code]